Theorem fneq1i 5059

Description: Equality inference for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq1i.1	|- F = G

Assertion

Ref	Expression
fneq1i	|- ( F Fn A <-> G Fn A )

Proof of Theorem fneq1i

Step	Hyp	Ref	Expression
1		fneq1i.1	. 2 |- F = G
2		fneq1 5053	. 2 |- ( F = G -> ( F Fn A <-> G Fn A ) )
3	1, 2	ax-mp 7	1 |- ( F Fn A <-> G Fn A )

Syntax hints:   <-> wb 103   = wceq 1285   Fn wfn 4962

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-3an 922  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2614  df-un 2988  df-in 2990  df-ss 2997  df-sn 3428  df-pr 3429  df-op 3431  df-br 3812  df-opab 3866  df-rel 4406  df-cnv 4407  df-co 4408  df-dm 4409  df-fun 4969  df-fn 4970

This theorem is referenced by:  fnunsn  5072  fnopabg  5088  f1oun  5219  f1oi  5237  f1osn  5239  ovid  5694  tfri1d  6030  frec2uzrand  9699  frec2uzf1od  9700  frecfzennn  9720